Prove that $y = \frac{4 \sin \theta}{2 + \cos \theta} - \theta$ is an increasing function of $\theta$ in the interval $\left[0, \frac{\pi}{2}\right].$

(A) Given function: $y = \frac{4 \sin \theta}{2 + \cos \theta} - \theta$
Find the derivative with respect to $\theta$ using the quotient rule:
$\frac{dy}{d\theta} = \frac{(2 + \cos \theta)(4 \cos \theta) - (4 \sin \theta)(-\sin \theta)}{(2 + \cos \theta)^2} - 1$
Simplify the expression:
$\frac{dy}{d\theta} = \frac{8 \cos \theta + 4 \cos^2 \theta + 4 \sin^2 \theta}{(2 + \cos \theta)^2} - 1$
Since $\cos^2 \theta + \sin^2 \theta = 1$:
$\frac{dy}{d\theta} = \frac{8 \cos \theta + 4(1)}{(2 + \cos \theta)^2} - 1 = \frac{8 \cos \theta + 4 - (4 + 4 \cos \theta + \cos^2 \theta)}{(2 + \cos \theta)^2}$
$\frac{dy}{d\theta} = \frac{4 \cos \theta - \cos^2 \theta}{(2 + \cos \theta)^2} = \frac{\cos \theta (4 - \cos \theta)}{(2 + \cos \theta)^2}$
In the interval $\theta \in \left[0, \frac{\pi}{2}\right]$,we know that $\cos \theta \ge 0$ and $4 - \cos \theta > 0$ (since $\cos \theta \le 1$).
Also,$(2 + \cos \theta)^2 > 0$ for all $\theta$.
Therefore,$\frac{dy}{d\theta} \ge 0$ for all $\theta \in \left[0, \frac{\pi}{2}\right]$.
Since the derivative is non-negative and the function is continuous on the closed interval,$y$ is an increasing function in $\left[0, \frac{\pi}{2}\right]$.